Reasoning: My guess would have to be that it has to do with the strength of the intermolecular forces between each molecule. For example, molecular compounds that have a very low molar mass like $\ce{CO}$ and $\ce{CH4}$. Both of theses molecules have a very small molar mass, a very low London dispersion force, and thus a very low boiling point. This in turns makes it easier to break the bonds that hold each individual atom and more than likely cause them to turn into a gas in normal conditions.

Then when you look at ionic compounds like $\ce{MgCl2}$ and $\ce{NaCl}$, it is more likely for these compounds to become solid because of the strong intermolecular force that they have (ionic bonds).

Is my reasoning correct and is there something else that I should consider?

  • $\begingroup$ Are you specifically interested in crystalline solids or just solids in general? The title implies the former but there doesn't seem to be any mention of it in the question. $\endgroup$
    – bon
    Commented May 22, 2015 at 17:22
  • $\begingroup$ @bon Reading my entry again, it would seem that you're right. I will change it to solids because is what I initially had in mind. $\endgroup$ Commented May 22, 2015 at 17:32

1 Answer 1


You are pretty much on the mark with looking at the strength of intermolecular forces. If you wanted to make a general comparison, the tendency to form a solid can be arranged by the strength of the intermolecular forces.$$\text{London Forces}<\text{Dipole-Dipole}<\text{Hydrogen bonding}<\text{Metallic}<\text{Ionic}<\text{Network Covalent}$$ Then, to compare within a group, you need some way of measuring the strength of that type of intermolecular force.

A more mathematical way to look at this if you have empirical values for a compound is to look at the $\Delta G_\text{fusion}$ at room temperature. $\Delta G_\text{fusion}=\Delta H_\text{fusion}-T\Delta S_\text{fusion}$, where $\Delta H_\text{fusion}$ is negative (since it will free up energy in forming the solid) and $\Delta S_\text{fusion}$ is negative (since, whether it started as a gas or a liquid, forming a solid will reduce entropy). Letting $\Delta G_\text{fusion}=0$ (reaction at equilibrium), we find that $T_{\text{fusion}}=\frac{\Delta H_\text{fusion}}{\Delta S_\text{fusion}}$. This allows us to see that when a compound becomes solid is a trade-off between the enthalpy released from forming bonds and the energy needed to reduce entropy at a given temperature.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.